Suppose a container is evacuated to leave just one molecule of a gas in it. Let v_(a) and v_(rms) represents the average speedand the rms speed of the gas.

Options

A. (a)`v_(a)gtv_(rms)`
B. (b)`v_(a)ltv_(rms)`
C. (c)`v_(a)=v_(rms)`
D. (d)`v_(rms)` is underfined.

Answer

To solve the problem, we need to analyze the relationship between the average speed (v_a) and the root mean square speed (v_rms) of a single molecule of gas in a container.

### Step-by-Step Solution:

1. **Understanding Average Speed (v_a)**:
- The average speed of a gas molecule is defined as the total distance traveled by the molecule divided by the total time taken.
- In the case of a single molecule, if it has a constant speed, then the average speed will simply be equal to that constant speed.

2. **Understanding Root Mean Square Speed (v_rms)**:
- The root mean square speed is defined as the square root of the average of the squares of the speeds of the gas molecules.
- For a single molecule with speed \( v \), the root mean square speed is calculated as:
\[ v_{rms} = \sqrt{\frac{v^2}{1}} = v \]
- Thus, for one molecule, the root mean square speed is equal to the speed of that molecule.

3. **Comparing Average Speed and RMS Speed**:
- Since both the average speed (v_a) and the root mean square speed (v_rms) are equal to the speed of the single molecule, we can conclude:
\[ v_a = v = v_{rms} \]

4. **Conclusion**:
- Therefore, for a container with only one molecule of gas, the average speed is equal to the root mean square speed:
\[ v_a = v_{rms} \]

### Final Answer:
The average speed \( v_a \) is equal to the root mean square speed \( v_{rms} \) when there is only one molecule of gas in the container.